I have to find the second prime derivative.
I got the first one, which resulted in the function in the above, but i have trouble doing the second derivative.

in the start i got....

From then on, i don't really know what to do. Can someone lecture me.

Jun 24th 2008, 03:24 PM

Krizalid

There's no need to apply the quotient rule here, first let's make this thing up:

Now differerentiate.

Jun 24th 2008, 03:28 PM

snakeman11689

Quote:

Originally Posted by Krizalid

There's no need to apply the quotient rule here, first let's make this thing up:

Now differerentiate.

What if i had to use the quotient rule...

Jun 24th 2008, 04:06 PM

Reckoner

Quote:

Originally Posted by snakeman11689

What if i had to use the quotient rule...

Krizalid has a much more elegant solution (though he made a little mistake), but if you insist:

(Factoring)

(Reducing and expanding)

Which is the same result that you will get through Krizalid's method (ignoring the mistake he made).

Quote:

Originally Posted by Krizalid

There's no need to apply the quotient rule here, first let's make this thing up:

Now differerentiate.

Nice work, but a tiny problem:

Have you spent so much time killing integrals that you've gone soft on the derivatives? :)

Jun 24th 2008, 04:17 PM

Krizalid

Yeah, it was a slight typo. :D

Jun 24th 2008, 04:43 PM

snakeman11689

Eventhough, I didn't understand what Krizalid did with my problem, I thank him for his contribution to my problem, so Thank You!!!

Thanks to Reckoner, I think i understand a bit more of the complex derivative that i did, It helped so much, thanks a bunch.

Jun 24th 2008, 05:24 PM

Reckoner

Quote:

Originally Posted by snakeman11689

Eventhough, I didn't understand what Krizalid did with my problem, I thank him for his contribution to my problem, so Thank You!!!

Krizalid used a technique called completing the square. You should have learned it in algebra, and if you have forgot it, I suggest you review as it is useful in many situations (especially with integration).

The idea is this: suppose you have some expression of the form . This type of expression can be inconvenient since you cannot directly combine the two terms due to the different exponents. So we add a constant value to the expression in such a way that we can factor it into the form . This is useful, for example, in solving quadratic equations (and indeed, this is where the quadratic formula comes from; see my derivation below).

Here is how to complete the square. Take your two terms (ignore anything else in the expression) and factor out the to get the expression into the form . Now, our goal is to get this in the form for some constant . If you expand, we get . We already have the , and setting gives the linear term. Thus, we only need to add . But if we add it, we must also subtract it so that the expression remains the same:

So, to summarize, these are the steps to complete the square:

* Start with some expression , where can be any real number. If the has a coefficient other than 1, factor it out first.

* Add and subtract the square of half of :

* Factor the first three terms:

*Done!

As an example, here is how you can use the method of completing the square to derive the quadratic formula: